Biological Computing

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Biological Computing DNA solution
Presented by Wooyoung Kim 4/8/09 CSc 8530
Parallel Algorithms, Spring 2009 Dr. Sushil K.
Prasad
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Outline

• NP and NP-complete
• Biological computation
• Hamiltonian path problem (HPP)
• Satisfaction problem
• Generalized SAT
• Discussion

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NP and NP-complete

• NP vs. NP-complete
• NP problems Non-deterministic Polynomial Time
  complexity.
• NP-complete all NP problems can be reduced to
  it, and if it has an efficient solution, then so
  do all NP problems.
• No general efficient solution exists for any
  NP-complete problem.

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Biological computation Adv.

• Speed of any computer is determined by
• How many parallel processes it has.
• How many steps each can perform per unit time.
• Biological computations could potentially have
  vastly more parallelism.
• Ex 3 g water contains approx. 1022 molecules.
• The second factor favors conventional computers,
  since biological machine is limited to small
  fraction of a biological experiment.
• However, the advantage in parallelism is so huge,
  the difference in the execution time is not a
  problem.

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Biological computation Disadv.

• Even with parallelism, brute force approach is
  not always feasible, too inefficient.
• The biological computer can solve any HPP of 70
  or less edges.
• Practically, there is not a great need, though.

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Hamiltonian Path Problem

• L.M. Adleman. "Molecular Computation of Solutions
  To Combinatorial Problem," Science, vol. 266,
  1994, pp 1021-1024.
• Using DNA, solve Hamiltonian Path Problem
  efficiently.

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Hamiltonian Path Problem
0 ? 1 ? 2 ? 3 ? 4 ? 5 ? 6
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Algorithm for HPP

1. Generating random paths through the graph.
2. Keep only those paths that begin with vin and end
   with vout.
3. If the graph has n vertices, then keep only those
   paths that enter exactly n vertices.
4. Keep only those paths that enter all of the
   vertices of the graph at least once.
5. If any paths remain, say Yes otherwise say
   No.

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Implementing Step 1

• Generating random paths through the graph.
• Ligation reaction (annealing)
• Each vertex encoded by random 20bp sequences (Oi)
• Approximately 3x1013 copies of the associated
  oligonucleotides (a short nucleic acid polymer)
  were added.

Vertex 2 (O2)
Vertex 3 (O3)
TATCGGATCG GTATATCCGA
GCTATTCGAG CTTAAAGCTA
GTATATCCGA GCTATTCGAG
Edge 2-gt3
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Implementing Step 2

• Keep only those paths that begin with vin (O0)and
  end with vout(O6).
• The product of step 1 were amplified by PCR
  (polymerase chain reaction) using O0(starting
  point) and O6(ending point)
• Thus keep only those molecules encode paths which
  begin with vin and end with vout.

O0
O6
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Implementing Step 3

• If the graph has n vertices, then keep only those
  paths that enter exactly n vertices.
• The product of Step2 was run on an agarose gel.
• The 140bp band (corresponding to double strand
  (ds) DNA encoding paths entering exactly seen
  vertices) was excised and soaked in ddH2O to
  extract DNA.

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Implementing Step 4

• Keep only those paths that enter all of the
  vertices of the graph at least once.
• The product of step 3 was affinity-purified with
  a biotin-avidin magnetic bead system, by
• First generating single stranded (ss) DNA from
  the dsDNA of step3
• Then incubating the ssDNA with the O1 conjugated
  to magnetic beads.
• Only those ssDNA containing O1 annealed to the
  bound O1, and were were retained.
• It is repeated with O2 until O5

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Implementing Step 5

• If any paths remain, say Yes otherwise say
  No.
• The product of step 4 was amplified by PCR and
  run on a gel.

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Drawbacks

• 7 days of lab work.
• Step 4 (magnetic bead separation) is most
  labor-intensive work.
• Possibility of errors
• Pseudo-paths
• Inexact reactions
• Hairpin loops

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Advantages

• The number of different oligonucleotides required
  should grow linearly with the number of edges.
• O(n)
• The fastest supercomputer vs. DNA computer
• 106 op/sec vs. 1014 op/sec
• 109 op/J vs. 1019 op/J (in ligation step)
• 1bit per 1012 nm3 vs. 1 bit per 1 nm3 (video
  tape vs. molecules)

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Satisfaction problem

• SAT consists of a Boolean formula of ,
  , where each Cl is a
  clause of the form .
  Vi is a variable or its negation. Ex.
• Problem find values of the variables so that
  the formula is 1.
• If we have n variables, then there are 2n choices
  to search.

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Satisfaction problem

• Graph formulation

unprimed ?1
primed ?0

• Suppose we have n variables in the formula,
  where ai represents the variables.
• This graph is constructed so that all paths from
  a1 to an1 encode an n-bit binary number.
• At each stage, a path has exactly two choices
  unprimed?1, primed?0
• Ex. A path a1x a2ya3 ? 01 , that is, x is 0 and
  y is 1.

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Satisfaction problem

• Example
• Number of variables n2 (x and y)
• Number of clauses m 2
• Construct a graph with (n1) 2n nodes for
  each clause and connect them as the following

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Satisfaction problem

• Graph paths and SAT problem

• If we have a path from a1 to an1 , that means
  each variable is represented by 0 or 1 and the
  formula satisfies.
• If there is no path from start to end, then the
  formula does not have any solution (not
  satisfies).
• Using the properties of DNA annealing
  (Watson-Crick complement binding), we can
  construct a graph representing the variables, and
  using test tubes, we can either obtain paths
  (satisfies) or no paths at all (not satisfies).

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Satisfaction problem

• Assign random pattern of DNA strings to each
  vertex. (ex. length 8)
• Then decide the pattern of DNA strings of each
  edge.

TATCCCGA
GGCTCGTT
GCAACCTA
CCTTATAG
GGCTAATG
CCCACCGA
ATTCGGAA
TTACGGGT
GGATTCCA
CCCAGGGT
TAATCCTA
CCTTCGAT
TCGAAATG
CCCAATTA
GCTAAGCT
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Satisfaction problem

• In an initial test tube t0, put many copies of
  the DNA strings corresponding the vertices and
  the edges. (many copies of each vertex and each
  edge)
• Put a sequence of complement of the first half
  of a1 and complement of the last half of a3 To
  show the start and end strings.

TATCCCGA
GGCTCGTT
TAAG
AGGT
ATTCGGAA
TTACGGGT
GGATTCCA
CCCAATTA
GCTAAGCT
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Satisfaction problem

1. Let t0 be an initial test tube containing all the
   DNA strings of vertices and edges.
2. Since the first clause is
   (that is, the first variable x is 1), operate
   E(t0,1,1) setting the first variable x to 1. Then
   extract only those corresponding patterns (10,11)
   and put it to t0-1
3. Put the remainder (pattern 00, 01), to t0-1 and
   operate E(t0-1,2,1) setting the second variable
   y to 1. Then extract only those corresponding
   patterns from t0-1 and put them to t0-2
4. Pour t0-1 and t0-2 together to form t1 test
   tubes.
5. Note that now the patterns of t1 is 01,10,11 and
   that is the solution of the first clause.

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Satisfaction problem

1. Repeat the same process for the second clause
   starting from t1.
2. Since the second clause is
   operate E(t1, 1, 0) to extract it to the t1-1
   test tube.
3. Put the remainder to t1-1 and make t1-2 by
   operating E(t1-1, 2,0).
4. Pour t1-1 and t1-2 into t2 test tube.
5. Check to see if there is any DNA in the last
   tube.
6. The satisfying assignments are exactly those in
   this final test tube.

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Satisfaction problem

Test tube OP Values
t0 initial 00, 01, 10, 11
t0-1 E(t0,1,1) 10, 11
t0-1 Reminder of t0-1 00, 01
t0-2 E(t0-1,2,1) 01
t1 Put t0-1 and t0-2 together 01, 10, 11
t1-1 E(t1, 1, 0) 01
t1-1 Reminder of t1-1 10,11
t1-2 E(t1-1,2,0) 10
t2 Put t1-1 and t1-2 together 01, 10
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Satisfaction problem

• For general formula with n variables and m
  clauses, we only need O(m) number of test tubes.
  (For each clause there are constant number of
  test tubes are additionally constructed)
• The last tube are checked to see if there is any
  patterns (paths) left from the start vertex to
  the end vertex.

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Generalized SAT

• Generalize this to consider problems that
  correspond to any Boolean formula.
• Formulas are defined by the recursive definition
• Any variable x is a formula
• If F is a formula, then so is F
• If F and G are formulas, then so are
  and

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Generalized SAT

• Size of the formula S the number of operations
  used to build the formula.
• SAT problem given a formula, find an assignment
  of Boolean values of variables so that the
  formula is true. ? NP-complete.
• Claim A O(S) number of DNA experiments can solve
  this SAT problem.

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Generalized SAT step1

• Construct a contact network for a formula.
• A contact network is a directed graph with source
  s and sink t
• Each edge is x or
• Given any assignment, an edge is connected if it
  is 1.

For example, the above graph is 1 only if w1 or
xyz1
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Generalized SAT step2

• Solve the SAT problem of a contact network by
  deciding
• Whether or not there is an assignment of values
  to the variables such that there is a directed
  connected path from s to t.
• If two edges have the same label, they should be
  consistent.
• How many of DNA experiments? O(S)

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Generalized SAT claims

• Note that the result follows from the two
  claims
• Given any formula of size S, there is a contact
  network of size linear in S , s.t. if the
  formula satisfies then the network satisfies.
• Given any contact network of size S, the SAT
  problem for the network can be solved in O(S) DNA
  experiments.

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Generalized SAT claim 1
Existence of contact network for given formula
simple formula Any formula can be placed into a
normal form with DeMorgans laws.

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Generalized SAT claim 1
Existence of contact network for given formula
general formula
G is a network for E, H is a network for F.

• The networks for
• (B) The networks for

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Generalized SAT claim2

• Solve the SAT problem for any contact network
  using O(S) number of DNA experiments
• Associate a test tube Pv with each node v in the
  contact network.
• The test tube Pt associated with the sink t is
  the answer
• Suppose that v?u is an edge with the label x and
  that Pv is already constructed. Then construct Pu
  by doing the extraction E(Pv, x,1)
• If several edges leave a vertex v then use an
  amplify step to get multiple copies in test tube
  Pv
• If several enter a vertex v, then pour the
  resulting test tubes together to form Pv.

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Discussion

• Can we actually build DNA computers?
• All the methods described here assumes that all
  the operations are perfect without error.
• However, the operations are not perfect.
• In the future, the DNA-based computers are hoped
  to be a practical means of solving hard problems.

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Reference

• R.J. Lipton. DNA solution of hard computational
  problems, Science, vol. 268, 1995, pp.542-545.
• L.M. Adleman. "Molecular Computation of Solutions
  To Combinatorial Problem," Science, vol. 266,
  1994, pp 1021-1024.
• R.J. Lipton. Speeding Up Computations via
  Molecular Biology, unpublished manuscript,
  available at www.cs.princeton.edu/rjl/